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Proof step 1 from proof min-ls-both (conclusion)

⊦ min(m, n) < m → (min(m, n) < n ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
min(m, n) < m, min(m, n) < n ⊦ m = 0, n = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1